poj_2823(單調隊列)

標籤：c++   基礎   演算法   單調隊列   

Sliding Window

Time Limit: 12000MS		Memory Limit: 65536K
Total Submissions: 38520		Accepted: 11415
Case Time Limit: 5000MS

Description

An array of size n ≤ 106 is given to you. There is a sliding window of sizek which is moving from the very left of the array to the very right. You can only see thek numbers in the window. Each time the sliding window moves rightwards by one position. Following is an example:
The array is [1 3 -1 -3 5 3 6 7], and k is 3.

Window position	Minimum value	Maximum value
[1  3  -1] -3  5  3  6  7	-1	3
1 [3  -1  -3] 5  3  6  7	-3	3
1  3 [-1  -3  5] 3  6  7	-3	5
1  3  -1 [-3  5  3] 6  7	-3	5
1  3  -1  -3 [5  3  6] 7	3	6
1  3  -1  -3  5 [3  6  7]	3	7

Your task is to determine the maximum and minimum values in the sliding window at each position.

Input

The input consists of two lines. The first line contains two integersn and k which are the lengths of the array and the sliding window. There aren integers in the second line.

Output

There are two lines in the output. The first line gives the minimum values in the window at each position, from left to right, respectively. The second line gives the maximum values.

Sample Input

8 31 3 -1 -3 5 3 6 7

Sample Output

-1 -3 -3 -3 3 33 3 5 5 6 7

 

一道裸的單調隊列，求區間最值問題。線段樹8秒多過，單調隊列4秒多過。可作為單調隊列的學習題目。

先說一下單調隊列是一種什麼樣的隊列。單調隊列分為遞增和遞減隊列，下面以遞增隊列為例：

1.該隊列時刻保持隊頭元素最小且遞增。元素從隊尾插入，如果小於隊尾元素，那麼就刪去隊尾元素，再插入。

2.同時需要一個數組儲存元素下標，以本題為例，區間長度為k,隊頭元素的下標小於當前元素下標i-k+1時，也就是此時隊頭元素已經不在k區間內了，就把隊頭元素刪去。

 

#include<iostream>#include<cstdio>#define M 1000001using namespace std;int n,k;int a[M];int q[M];int p[M];void get_min(){    int head=1;    int tail=0;    for(int i=0;i<k-1;i++)    {        while(head<=tail&&a[i]<=q[tail])            --tail;        q[++tail]=a[i];        p[tail]=i;    }    for(int i=k-1;i<n;i++)    {        while(head<=tail&&a[i]<=q[tail])            --tail;        q[++tail]=a[i];        p[tail]=i;        while(p[head]<i-k+1)        {            ++head;        }        cout<<q[head]<<" ";    }    cout<<endl;}void get_max(){    int head=1;    int tail=0;    for(int i=0;i<k-1;i++)    {        while(head<=tail&&a[i]>=q[tail])            --tail;        q[++tail]=a[i];        p[tail]=i;    }    for(int i=k-1;i<n;i++)    {        while(head<=tail&&a[i]>=q[tail])            --tail;        q[++tail]=a[i];        p[tail]=i;        while(p[head]<i-k+1)        {            ++head;        }        cout<<q[head]<<" ";    }    cout<<endl;}int main(){    //freopen("d:\\test.txt","r",stdin);    scanf("%d%d",&n,&k);    for(int i=0;i<n;i++)    {        scanf("%d",&a[i]);    }    get_min();    get_max();    return 0;}